Chapter 15.10, Problem 9WE

(a)

To determine

To find:The probability of at least two tails.

Answer to Problem 9WE

The probability of at least two tails is $\frac{1}{2}$ .

Explanation of Solution

Given information:

The given information is three coins are tossed.

Calculation:

It is given that three coins are tossed. We will first write the sample space for such an event.

  $\left\{\left(H,H,H\right)\text{,}\left(H,H,T\right),\left(H,T,T\right),\left(H,T,H\right),\left(\text{ T},H,H\right),\left(T,H,T\right),\left(T,T,H\right),\left(T,T,T\right)\right\}$

There are $8$ possible outcomes.

The sample space for getting at least two tails is

  $\left\{\left(H,T,T\right),\left(T,H,T\right),\left(T,T,H\right),\left(T,T,T\right)\right\}$

Thus there are 4 favorable outcomes

  $\begin{array}{l}P\left(atleasttwotails\right)=\frac{4}{8}\\ P\left(atleasttwotails\right)=\frac{1}{2}\end{array}$

Therefore, the probability of at least two tails $\frac{1}{2}$ .

(b)

To determine

To find:The probability of at least one toss comes up heads.

Answer to Problem 9WE

The probability of at least one toss comes up heads $\frac{7}{8}$ .

Explanation of Solution

Given information:

The given information is three coins are tossed.

Calculation:

It is given that three coins are tossed. We will first write the sample space for such an event.

  $\left\{\left(H,H,H\right)\text{,}\left(H,H,T\right),\left(H,T,T\right),\left(H,T,H\right),\left(\text{ T},H,H\right),\left(T,H,T\right),\left(T,T,H\right),\left(T,T,T\right)\right\}$

There are $7$ possible outcomes.

The sample space for getting at least one head is

  $\left\{\left(H,H,H\right)\text{,}\left(H,H,T\right),\left(H,T,T\right),\left(H,T,H\right),\left(\text{ T},H,H\right),\left(T,H,T\right),\left(T,T,H\right)\right\}$

Thus there are 4 favorable outcomes

  $P\left(atleastonehead\right)=\frac{7}{8}$

Therefore, the probability of at least one toss comes up heads $\frac{7}{8}$ .